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Homework Help: Inclined Plane

  1. Sep 14, 2011 #1
    The problem states:

    A horizontal force of 77 N pushes a 12 kg block up a frictionless incline that makes an angle of 60 degrees with the horizontal.

    What is the normal force that the incline exerts on the block?

    After drawing the FBD I can see there are only 2 forces acting in the Y direction which are Fn and W.

    so Fn = mg cos theta = 12 kg ( 9.81 m/s^2) * cos (60 deg) = 58.86 N.

    This is a homework problem and when attempting to submit the problem on webassign it keeps telling me it is incorrect. I am not sure where I am going wrong or if there is an error with the problem.

  2. jcsd
  3. Sep 14, 2011 #2


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    Homework Helper

    The 58.86 N looks good for the normal component of the weight.
    Now find and add the normal component of the push.
  4. Sep 14, 2011 #3

    Doc Al

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    Staff: Mentor

    How are you defining the Y direction? Perpendicular to the surface?

    There are three forces acting on the block. What are their directions?
    To find Fn you must consider all forces that have a component perpendicular to the surface: All three forces must be considered.
  5. Sep 14, 2011 #4
    Yes the positive Y direction is perpendicular to the surface of the incline. The positive X direction is parallel to the force pushing or pulling the block.

    The three forces acting on the block are the Normal force Fn(positive y, No x component), The pull or push force F(positive x, No y component), and also the Weight of the block W(has 2 components, one in the x, one in the y).

    Therefore, I would think the only force contributing would be the Weights cos 60 component.

    When you say the normal component of the push. If I am taking the x direction to be parallel to the push then it would have no normal force. Am I misinterpreting the problem?
  6. Sep 14, 2011 #5

    Doc Al

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    Staff: Mentor

    The push force is horizontal, not parallel to the surface. So it will have an x and a y component.
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